Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $10$ years; the standard deviation is $1.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $7.2$ years.
Explanation: $10$ $8.6$ $11.4$ $7.2$ $12.8$ $5.8$ $14.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $10$ years. We know the standard deviation is $1.4$ years, so one standard deviation below the mean is $8.6$ years and one standard deviation above the mean is $11.4$ years. Two standard deviations below the mean is $7.2$ years and two standard deviations above the mean is $12.8$ years. Three standard deviations below the mean is $5.8$ years and three standard deviations above the mean is $14.2$ years. We are interested in the probability of a lion living less than $7.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $7.2$ years and the other half $({2.5\%})$ will live longer than $12.8$ years. The probability of a particular lion living less than $7.2$ years is ${2.5\%}$.